# How do you find the distance between point A(-3, 5) and point B(4, -6) in the coordinate plane?

Jun 10, 2015

$d \left(A , B\right) = \sqrt{170} \approx 13.03$.

#### Explanation:

Having the points $A \left(- 3 , 5\right) \mathmr{and} B \left(4 , - 6\right)$ on the same plane.
We calculate the distance d with the formula:
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$.
So in this case we have:
$d = \sqrt{{\left(4 - \left(- 3\right)\right)}^{2} + {\left(- 6 - 5\right)}^{2}} = \sqrt{49 + 121} = \sqrt{170} \approx 13.03$

Jun 10, 2015

You first work out the $x$- and $y$-distances and then use Pythagoras

#### Explanation:

In the horizontal (x) direction the distance is:
$\Delta x = 4 - \left(- 3\right) = 7$
In the vertical (y) direction the distance is:
$\Delta y = - 6 - 5 = - 11$ or just plain $11$
(you should draw this on coordinate paper)

We now have a right-sided triangle, where the hypotenuse is the distance between A and B.

So ${\left(A B\right)}^{2} = {\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} \to$
${\left(A B\right)}^{2} = {7}^{2} + {11}^{2} = 49 + 121 = 170 \to$
$A B = \sqrt{170} \approx 13.0$

Jun 10, 2015

Use formula for distance: ${d}^{2} = {\left(x 2 - x 1\right)}^{2} + {\left(y 2 - y 1\right)}^{2}$
${d}^{2} = {\left(4 + 3\right)}^{2} + {\left(- 6 - 5\right)}^{2} = 49 + 121 = 170$